Statement 1: An electric motor will have maximum efficiency when back emf becomes equal to half of applied emf.

Statement 2: Efficiency of electric motor depends only on magnitude of back emf.

Since the efficiency of an electric motor is given by

$\eta =\frac{\text{ output power }}{\text{ input power }}$

From the above relation, it is quite clear that maximum output power corresponds maximum efficiency of motor.

Now, output power is given by

$\mathrm{eI}=\frac{\mathrm{e}(\mathrm{E}-\mathrm{e})}{\mathrm{R}}$                 …(i)

To obtain maximum output power, differentiating Eq. 
(i) with respect to e which will be equal to zero.

So,  $\frac{\mathrm{d}}{\mathrm{de}}\left[\frac{\mathrm{e}(\mathrm{E}-\mathrm{e})}{\mathrm{R}}\right]=0\Rightarrow \mathrm{e}=\frac{\mathrm{E}}{2}$

Thus, when back emf becomes equal to half of the applied emf, the efficiency of motor will be maximum.

$Also \eta =\frac{ei}{Ei}=\frac{e}{E} .So efficiency depends on back emf and supply voltage$.